Coupled Oscillators - lmao PDF

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Stephen Natanael Basatua U2040050A Experiment 2: Coupled Oscillators

Aim To find the normal modes of different number of oscillating bodies and compare them to theoretically calculated value.

Theoretical Background 1 Mass 2 Springs Consider a mass with springs attached

Figure 1 2 Springs 1 Mass System

This system’s normal mode is

ω=

√

2k (1) m

and its maximum amplitude at resonance (considering friction) is

A=

F0 2

4π m f0

(2)

2 Masses 3 Springs Consider 2 masses with springs attached

Figure 2 3 Springs 2 Masses System

Using Newtonian mechanics, we obtain the following equation with 2 degree of freedom

x 1 ( −2 k + m ω2 )+ x 2 ( k )=0(3) x 1 (k ) + x 2 (−2 k +m ω2 )=0(4) We will find its normal modes by solving them as matrices

(

−2 k +m ω k

While

|

|

2

−2 k +m ω k 2 =0 k −2 k +m ω 4

2

2

2

ω m −4 km ω +3 k =0

2

)( )

x1 k =0 ( 5 ) −2 k +m ω2 x2

Stephen Natanael Basatua U2040050A 2

ω1,2=

2k ±k m

Eventually we have 2 normal modes for this system: ω1 =

√

3k m

and

√

ω2 =

k m

Apparatus 1.Mechanical Vibrator 2.Motion Sensor 3.Pulley 4.Glider with flag 5.Spring 6.Mass and Hanger set 7.Universal 850 Interface

Procedures 1.Determine the spring constants of the springs by using the stand, mass hangers and masses provided. 2.Measure mass of the gliding masses and estimate the theoretical resonant frequency. 3.Turn on the air supply and maintain optimal flow rate indicated by the arrow near the dial. 4.Set up the apparatus. Follow the precautions. 5.Adjust the adjustable end stop until the springs are stretched to about 15 cm each. 6.Make sure the motion sensor is in line with the flag on top of the glider 7.Connect the motion sensors to the digital channels 1 to 4 of the 850 Interface. 8.Connect the leads from output 1 of USB850 to the mechanical vibrator. Output 1 will be the source of the AC signal driving the oscillations of the system studied. 9.Open CAPSTONE a. Click on “hardware setup” tab on the left. The 850 Interface should be automatically detected; a cartoon of the interface is visible with digital inputs. b. Right click on the digital inputs and select the “motion sensor” from the drop-down menu that you see. c. Click on “Signal generator” tab and select the “850 Output 1”. For this output i. Select sine wave of magnitude 0.3 V. ii. Frequency f must be less than 5Hz. 10.Check that the response of the motion sensors on the computer can be detected. 11.For different frequencies of not more than 5.0 Hz, plot the amplitude of the vibration of springs against driving frequency. Determine the resonant frequency (f ~ f0) and compare it against the theoretical result. 12.Repeat the experiment by adding another mass into the system, making it a three-spring and twomass system. 13.Calculate the theoretical normal mode frequencies of the three-spring, two-mass system.

Stephen Natanael Basatua U2040050A

Results 2 Springs 1 Mass

Graph 1Amplitude vs Driving frequency (1 mass 2 springs) We found the peak of this graph to be at 0.900Hz which is close to the theoretical value f 0=0.9 11 Hz . We can see that as the driving frequency approaches f 0 , the higher the amplitude becomes but limited due to friction 3 Springs 2 Masses Mass A

Graph 2 Amplitude vs Driving frequency of Mass A

Stephen Natanael Basatua U2040050A Similar case as Graph 1, when driving frequencies approaches any of the 2 natural frequencies, the amplitude goes higher but goes down every time it passes through any of the natural frequencies until it reaches a certain frequency which yields its lowest amplitude.

Mass B

Graph 3 Amplitude vs Driving frequency of Mass B We obtained the first normal mode to be 0.650 Hz and the second normal mode to be 1.150Hz. Theoretically, the values are 0.644 Hz and 1.120 Hz subsequently.

Error Analysis Firstly we assume all the springs used in this experiment have the same value k. The mass for each glider is m=199.3 g . The value for k is obtained from hanging a mass m w stationarily. While m w =50 g ( δm=0.05 g ¿ and the spring stretched Δx =15 cm (δx=0.05 cm) . So

k=

mw g (6)=3.26 7 N /m Δx m g δk= w Δx

√(

)( ) 2

k best =(3.26 7 ± 0.011)N /m The normal modes for any system can be defined as

ω=

√

where

δ ω=

βk (8) m β

√

is some constant. Obviously

β2 δ k 2 β 2 k δ m2 (9) + 4 km m3

2

δm δx + (7)=0.011 N /m mw Δx

Stephen Natanael Basatua U2040050A Note that the theoretical values for the natural frequencies of the system above are obtained from equation (8) divided by 2 pi whereas the experimental values are obtained from peak analysis from each graph For 2 springs 1 mass system we have

β=2 which yield

ω best=5.726 ± 0.019 Hz For 3 springs 2 masses system we have β=3

which yield

ω best=7.012 ±0.018 Hz and

β=1 which yield

ω best=4.049± 0.015 Hz From these results we can justify that there were obvious sources of error: 1.Each spring was assumed to be identical although they were not (systematic) 2.Spring constant calculation did not take account of the spring’s mass (systematic) 3.Some springs may have deformed slightly, resulting unexpected stretch length (systematic) 4.False reading (human error) 5.Refresh rate of the position sensor (systematic) Conclusion

ω2 springs 1 mass=5.726 ± 0.019 H z ,

ω3 springs2 masses β=3=7.012 ± 0.018 H z ,

ω3 springs2 masses β=1=4.049± 0.015 Hz Reference

1. Blackboard. (n.d.). NTU Learn. Retrieved February 18, 2021, from https://ntulearn.ntu.edu.sg/webapps/blackboard/execute/content/file? cmd=view&content_id=_2417766_1&course_id=_327044_1 2. Lecture3-Coupled-Oscillators. (n.d.). Retrieved from https://scholar.harvard.edu/files/schwartz/files/lecture3-coupled-oscillators.pdf

Stephen Natanael Basatua U2040050A...